Sinx=1/4 , x in 2nd quadrant... find sinx/2 , cosx/2, tanx/2
Answers
Since is in the 2nd quadrant, is in the first quadrant.
Given . Now
Since is in the 2nd quadrant,
Now use the identity,
Step-by-step explanation:
Since xx is in the 2nd quadrant, x/2x/2 is in the first quadrant.
Given \sin x=\frac{1}{4}sinx=
4
1
. Now
\begin{gathered}\cos x=\pm \sqrt{1-\sin^2 x} \\ \cos x=\pm \sqrt{1-\frac{1}{4}^2} \\ \cos x=\pm \frac{\sqrt{15}}{4}\end{gathered}
cosx=±
1−sin
2
x
cosx=±
1−
4
1
2
cosx=±
4
15
Since xx is in the 2nd quadrant, \cos x=- \frac{\sqrt{15}}{4} < 0cosx=−
4
15
<0
Now use the identity,
\begin{gathered}2\cos ^2(x/2)=1+\cos x\\ 2\cos ^2(x/2)=1- \frac{\sqrt{15}}{4} \\ \cos ^2(x/2)= \frac{4-\sqrt{15}}{8} \\ \cos (x/2)=\sqrt{ \frac{4-\sqrt{15}}{8} } \\ \sin (x/2)=\sqrt{1-\cos^2 (x/2)}\end{gathered}
2cos
2
(x/2)=1+cosx
2cos
2
(x/2)=1−
4
15
cos
2
(x/2)=
8
4−
15
cos(x/2)=
8
4−
15
sin(x/2)=
1−cos
2
(x/2)
\begin{gathered}\sin (x/2)=\sqrt{1-\frac{4-\sqrt{15}}{8} }\\ \sin (x/2)=\sqrt{ \frac{4+\sqrt{15}}{8} } \\ \tan (x/2)=\frac{\sin (x/2)}{\sin (x/2)} \\ \tan (x/2)=\frac{\sqrt{ \frac{4+\sqrt{15}}{8} } }{\sqrt{ \frac{4-\sqrt{15}}{8} } } \\ \tan (x/2)=\sqrt{\frac{4+\sqrt{15}}{4-\sqrt{15}}}\end{gathered}
sin(x/2)=
1−
8
4−
15
sin(x/2)=
8
4+
15
tan(x/2)=
sin(x/2)
sin(x/2)
tan(x/2)=
8
4−
15
8
4+
15
tan(x/2)=
4−
15
4+
15